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#1520 by SeeShells on 09 Jan, 2016 19:08
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In the above video, he mentions that as they were trying to improve Q, they started with copper, then went to niobium, and then went back to copper (but highly polished) and coated with a thin layer of niobium.
A sputtered layer of niobium ! (We do sputtering)
A clarification for the general audience and DIY's, concerning the sensitivity of the electrical conductivity (and hence the quality of resonance Q) to the temperature range :
QUESTION: What is the explanation for a higher quality of resonance (Q) with Niobium than with Copper ?
ANSWER: It has to do with the superconducting temperature range, but not at room temperature. At ambient conditions, copper is 9 times better than Niobium concerning electrical conductivity, and therefore concerning Quality of Resonance (Q).
The superconducting temperature of Niobium is 9.3 K (that's significantly below the boiling point at room pressure of Liquid Nitrogen: 77 K, and below the critical point of Liquid Hydrogen: 33 K)
Copper is not a superconductor. The best conductors at room temperature (gold, silver, and copper) do not become superconducting at all. They have the smallest lattice vibrations, therefore the electron phonon interaction is weak in copper, silver or gold so that Coulomb effects win and prevent superconductivity.
See the commonly quoted electrical resistivity and conductivities of these materials at ambient conditions:
________________
According to
1) https://en.wikipedia.org/wiki/Niobium
Niobium Electrical resistivity = 1.52 ×10^(−7) Ω•m ( at 0 °C)
therefore
Niobium Electrical conductivity = 6.579 *10^6 S/m
2) http://www.webelements.com/niobium/physics.html
Niobium Electrical resistivity = 1.52 ×10^(−7) Ω•m
therefore
Niobium Electrical conductivity = 6.579 *10^6 S/m
3) http://www.allmeasures.com/Formulae/static/materials/140/electrical_resistivity.htm
Niobium Electrical resistivity = 1.44 ×10^(−7) Ω•m
therefore
Niobium Electrical conductivity = 6.944 *10^6 S/m
________________
4) http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/
Pure Copper Electrical resistivity = 1.664 ×10^(−
Ω•m
therefore
Pure Copper Electrical conductivity = 6.009 *10^7 S/m
_________________
According to the above quoted data, Pure Copper has an Electrical Conductivity that is 9 times greater than Niobium at ambient temperatures, therefore everything else being equal, pure copper should give a Q 9 times better than Niobium, at ambient conditions (for polished and non-corroded copper)
_________________
It is at superconducting temperatures that Niobium's electrical conductivity is superior to pure copper.
See:
http://arxiv.org/ftp/cond-mat/papers/0308/0308266.pdf
I just got on and read what you researched. I came to the same conclusion this morning about Niobium and being a superconductor, better than copper.
Another thing struck me at the same time is when aero runs his meep simulations with the perfect conductor is he simulating superconducting materials as well? Is this why we saw a Q in one of those tests of over 11 billion?
Interesting or in this case Fascinating.
Shell
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#1521 by Rodal on 09 Jan, 2016 19:20
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...
I just got on and read what you researched. I came to the same conclusion this morning about Niobium and being a superconductor, better than copper.
Another thing struck me at the same time is when aero runs his meep simulations with the perfect conductor is he simulating superconducting materials as well? Is this why we saw a Q in one of those tests of over 11 billion?
Interesting or in this case Fascinating.
Shell
There was a units problem in previous Meep runs, as I recall of 113 times (going by memory).
If that error was involved, it could bring down the calculated Q from 11,000 million to about 100 million, which is still huge even by superconducting standards, so it appears to me that this 11,000 million Q value may be due to a numerical artifact in the calculation of Q.
This appears to be confirmed by the fact that the -11,000 million Q output by Meep has a negative value, which means that Meep was showing energy production instead of energy dissipation !!!
So, it looks like there was something very, very wrong with those Meep runs (outputting negative Q).
The highest positive value of Q in that list is 88 million, which when divided by 113 gives a Q of 780,000.
If you look at the classic book "Dielectric and Waves" by Prof. Arthur Von Hippel from MIT (who is famous for his work during WWII, and whose book is dedicated by Nobel Prize celebrities Niels Bohr and James Frank who were colleagues of Von Hippel), on page 81 it gives the following information from experiments at MIT during WWII at the MIT Radiation Lab (*):Quote from: Prof. Arthur Von HippelThe great advantage of cavity resonators is their excellent "ringing" quality in comparison with that of open-line resonators. Since no energy is lost by radiation, Q values of 10^4 can be obtained without much difficulty, and even 10^6 may be realized by the excitation of favorable modes in large cavities, whereas values not higher than 10^2 may be expected from an open transmission line in the microwave region
I stress that this quotation from Prof. Von Hippel dates from his book edition of 1954 based on experiments at MIT.
It was only after the publication of Von Hippel's book, in 1955, that George Yntema succeeded in constructing a small 0.7-tesla iron-core electromagnet with superconducting niobium wire windings. Then, in 1961, J.E. Kunzler, E. Buehler, F.S.L. Hsu, and J.H. Wernick made the startling discovery that, at 4.2 degrees kelvin, a compound consisting of three parts niobium and one part tin, was capable of supporting a current density of more than 100,000 amperes per square centimeter in a magnetic field of 8.8 tesla. Despite being brittle and difficult to fabricate, niobium-tin has since proved extremely useful in supermagnets generating magnetic fields as high as 20 tesla.
You will see that Von Hippel's quotation of a Q of 10,000 "without much difficulty" is similar to the values that rfmwguy wrote he was familiar with. Therefore it is interesting that Prof. Von Hippel wrote in 1954 that a Q of 1 million may be realized by "the excitation of favorable modes in large cavities"
Since there was only so much that one could do regarding the material (using silver or copper), it seems to me that the key word is "the excitation of favorable modes in large cavities", and that since the quality factor Q is proportional to the ratio (energyVolumeIntegral/energySurfaceIntegral) of the electromagnetic fields, obviously Prof. Von Hippel is referring to a mode in a large cavity that has a huge ratio of the Energy Volume Integral to the Energy Surface Integral
___________
(*) Together with MIT's Radiation Lab, von Hippel and his collaborators helped to develop radar technology during the war. He was awarded the President's Certificate of Merit in 1948 by U.S. President Harry Truman. He became famous also for his discovery of ferroelectric and piezoelectric properties of barium titanate (BaTiO3).
During the war the results on dielectrics obtained by his "Laboratory for Insulation Research" were classified information. After the war these results were prepared for publication. -
#1522 by X_RaY on 09 Jan, 2016 20:06
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That you can't solve this set of equations in the under cut off volume doesn't mean there can't be thrust at all.I agree with 1) but what is the evidence that leads to 2)?
Why does Shawyer suggest to keep the small end above cutoff?
It is stated in Roger's thrust equation:
Thrust = (2 * Qu * Df * Power) / c
Where Df is as attached and is where the small end cutoff comes into the equation. With the small end in cutoff, there is no thrust.
Please refer to the 2nd and 3rd attachments as well
(If there was ever thrust like discussed and not other effects like Lorentz force or whatever)
As far as I know Shawyer's Df and thrust equations are not proven till now.
All discussed thrust equations for conical cavities leads to larger thrust the closer the narrow end of the cavity is in relation to it's apex (mathematically singularities at null distance).
Some threads ago we had a huge discussion especially of the effects of evanescent waves.
Question: Was this been tested by Shawyer and his company or is it his intuitive opinion that it would not work with an undersized end of the cone, just below cutoff diameter?
Roger has told me and others that the small end MUST operate above cutoff or no thrust. Would expect that was from coal face experience plus there is no Df if the small end is at or below cutoff.
So his advise matches his equation.
I was hopeful to get more informations about than that statement (Formula, graphs, calculations ,experimental results or something else from SPR, ...) nevertheless thanks for your answer.
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#1523 by DaCunha on 09 Jan, 2016 20:10
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In order to reach even larger Q in a stable state it is important to have a temperature difference as puffer between nitrogen temp (77 K) and transition temp.
This is because the critical magnetic field strength which will make supraconductivity collapse is temperature dependent.
The lower under critical temp you are the higher will the energy be you can store inside the cavity before SC collapse.
Because it seems we have sputtering and physical vapour deposition experts here.
I ask: Would it be possible to organize a crowdfunding for one of the replicators here and let her/him coat her/his surface with Bi2Sr2Ca2Cu3O10 (110 K) or even Hg0,8Tl0,2Ba2Ca2Cu3O8 (138 K) ?
Just think of it, this would be the perfect way to exclude all thermal effects related systematic errors from the setup.
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#1524 by SeeShells on 09 Jan, 2016 20:25
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...
I just got on and read what you researched. I came to the same conclusion this morning about Niobium and being a superconductor, better than copper.
Another thing struck me at the same time is when aero runs his meep simulations with the perfect conductor is he simulating superconducting materials as well? Is this why we saw a Q in one of those tests of over 11 billion?
Interesting or in this case Fascinating.
Shell
There was a units problem in previous Meep runs, as I recall of 113 times (going by memory).
If that error was involved, it could bring down the calculated Q from 11,000 million to about 100 million, which is still huge even by superconducting standards, so it appears to me that this 11,000 million Q value may be due to a numerical artifact in the calculation of Q.
This appears to be confirmed by the fact that the -11,000 million Q output by Meep has a negative value, which means that Meep was showing energy production instead of energy dissipation !!!
So, it looks like there was something very, very wrong with those Meep runs (outputting negative Q).
The highest positive value of Q in that list is 88 million, which when divided by 113 gives a Q of 780,000.
If you look at the classic book "Dielectric and Waves" by Prof. Arthur Von Hippel from MIT (who is famous for his work during WWII, and whose book is dedicated by Nobel Prize celebrities Niels Bohr and James Frank who were colleagues of Von Hippel), on page 81 it gives the following information from experiments at MIT during WWII at the MIT Radiation Lab (*):Quote from: Prof. Arthur Von HippelThe great advantage of cavity resonators is their excellent "ringing" quality in comparison with that of open-line resonators. Since no energy is lost by radiation, Q values of 10^4 can be obtained without much difficulty, and even 10^6 may be realized by the excitation of favorable modes in large cavities, whereas values not higher than 10^2 may be expected from an open transmission line in the microwave region
I stress that this quotation from Prof. Von Hippel dates from his book edition of 1954 based on experiments at MIT.
It was only after the publication of Von Hippel's book, in 1955, that George Yntema succeeded in constructing a small 0.7-tesla iron-core electromagnet with superconducting niobium wire windings. Then, in 1961, J.E. Kunzler, E. Buehler, F.S.L. Hsu, and J.H. Wernick made the startling discovery that, at 4.2 degrees kelvin, a compound consisting of three parts niobium and one part tin, was capable of supporting a current density of more than 100,000 amperes per square centimeter in a magnetic field of 8.8 tesla. Despite being brittle and difficult to fabricate, niobium-tin has since proved extremely useful in supermagnets generating magnetic fields as high as 20 tesla.
You will see that Von Hippel's quotation of a Q of 10,000 "without much difficulty" is similar to the values that rfmwguy wrote he was familiar with. Therefore it is interesting that Prof. Von Hippel wrote in 1954 that a Q of 1 million may be realized by "the excitation of favorable modes in large cavities"
___________
(*) Together with MIT's Radiation Lab, von Hippel and his collaborators helped to develop radar technology during the war. He was awarded the President's Certificate of Merit in 1948 by U.S. President Harry Truman. He became famous also for his discovery of ferroelectric and piezoelectric properties of barium titanate (BaTiO3).
During the war the results on dielectrics obtained by his "Laboratory for Insulation Research" were classified information. After the war these results were prepared for publication.
Of course 11 Billion is not a real number by any means, but the real question I wanted answered is, can meep simulate a superconducting cavity with the current perfect metal settings or would we have to modify the drude model?
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#1525 by aero on 09 Jan, 2016 20:30
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...
I just got on and read what you researched. I came to the same conclusion this morning about Niobium and being a superconductor, better than copper.
Another thing struck me at the same time is when aero runs his meep simulations with the perfect conductor is he simulating superconducting materials as well? Is this why we saw a Q in one of those tests of over 11 billion?
Interesting or in this case Fascinating.
Shell
There was a units problem in previous Meep runs, as I recall of 113 times (going by memory).
If that error was involved, it could bring down the calculated Q from 11,000 million to about 100 million, which is still huge even by superconducting standards, so it appears to me that this 11,000 million Q value may be due to a numerical artifact in the calculation of Q.
This appears to be confirmed by the fact that the -11,000 million Q output by Meep has a negative value, which means that Meep was showing energy production instead of energy dissipation !!!
So, it looks like there was something very, very wrong with those Meep runs (outputting negative Q).
The highest positive value of Q in that list is 88 million, which when divided by 113 gives a Q of 780,000.
If you look at the classic book "Dielectric and Waves" by Prof. Arthur Von Hippel from MIT (who is famous for his work during WWII, and whose book is dedicated by Nobel Prize celebrities Niels Bohr and James Frank who were colleagues of Von Hippel), on page 81 it gives the following information from experiments at MIT during WWII at the MIT Radiation Lab (*):Quote from: Prof. Arthur Von HippelThe great advantage of cavity resonators is their excellent "ringing" quality in comparison with that of open-line resonators. Since no energy is lost by radiation, Q values of 10^4 can be obtained without much difficulty, and even 10^6 may be realized by the excitation of favorable modes in large cavities, whereas values not higher than 10^2 may be expected from an open transmission line in the microwave region
I stress that this quotation from Prof. Von Hippel dates from his book edition of 1954 based on experiments at MIT.
It was only after the publication of Von Hippel's book, in 1955, that George Yntema succeeded in constructing a small 0.7-tesla iron-core electromagnet with superconducting niobium wire windings. Then, in 1961, J.E. Kunzler, E. Buehler, F.S.L. Hsu, and J.H. Wernick made the startling discovery that, at 4.2 degrees kelvin, a compound consisting of three parts niobium and one part tin, was capable of supporting a current density of more than 100,000 amperes per square centimeter in a magnetic field of 8.8 tesla. Despite being brittle and difficult to fabricate, niobium-tin has since proved extremely useful in supermagnets generating magnetic fields as high as 20 tesla.
You will see that Von Hippel's quotation of a Q of 10,000 "without much difficulty" is similar to the values that rfmwguy wrote he was familiar with. Therefore it is interesting that Prof. Von Hippel wrote in 1954 that a Q of 1 million may be realized by "the excitation of favorable modes in large cavities"
___________
(*) Together with MIT's Radiation Lab, von Hippel and his collaborators helped to develop radar technology during the war. He was awarded the President's Certificate of Merit in 1948 by U.S. President Harry Truman. He became famous also for his discovery of ferroelectric and piezoelectric properties of barium titanate (BaTiO3).
During the war the results on dielectrics obtained by his "Laboratory for Insulation Research" were classified information. After the war these results were prepared for publication.
Actually Dr. Rodal, I believe that run, Q > 11 billion was with a perfect metal model. The broken copper model run gave a Q of "only" about 400 million.
As for the "minus" sign, that is a recurring confusion in meep resonance calculations. I believe it indicates that Harminv has not completely converged. I don't understand why it stops before it completely converges though, Harminv is very fast so a few more iterations wouldn't even be noticeable.
What would be the difference between a meep "perfect metal" material and an ideal superconductor? -
#1526 by Rodal on 09 Jan, 2016 20:34
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..
Of course 11 Billion is not a real number by any means, but the real question I wanted answered is, can meep simulate a superconducting cavity with the current perfect metal settings or would we have to modify the drude model?
1) The problem is not just that 11 billion Q is not a real number, the problem is understanding why Meep gave that output. The problem is understanding why those Meep runs gave a negative huge value of Q of 11 billion, which means that energy was being produced in the Meep run at an enormous rate. If that bug is not understood (garbage into Meep = garbage out of Meep) then one cannot really have confidence in any other Meep run whether at ambient temperature or superconducting. It is a numerical problem with the model.
2) Meep is capable of modeling the superconducting regime, if the material properties are input correctly and if the Finite DIfference space mesh is converged and the finite difference time step is small enough, and the matrix is not ill-conditioned etc.
3) Given aero's information HarmInv was involved. It may be due to numerical ill-conditioning. One would have to find out what was responsible for the numerical ill-conditioning. -
#1527 by DaCunha on 09 Jan, 2016 20:35
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It is definitely not that simple. See my post above.
A superconductor can lose it's state if external magnetic fields slowly oscillating fields are too strong or if photon energies of incident radiation is larger than the cooper binding energies.
If you manage to take this effects into account than the perfect metal approximation would be good. -
#1528 by Rodal on 09 Jan, 2016 20:41
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It is definitely not that simple. See my post above.
One would have to look exactly as to:
A superconductor can lose it's state if external magnetic fields slowly oscillating fields are too strong or if photon energies of incident radiation is larger than the cooper binding energies.
If you manage to take this effects into account than the perfect metal approximation would be good.
1) How was the Q calculated by Meep. What equation was used to calculate Q?
2) What is meant by a "perfect conductor" model in Meep? What are the exact equations used inside it to model a "perfect conductor" ? Obviously if it was "perfect", conductivity would be infinite and the Q would be infinite. So if the Q is finite, does that mean that the perfect conductor model, is not so perfect, and the "perfect conductivity" has a finite value ?
What is the sense of calculating a numerical value for Q for a perfect metal if the perfect metal is modeled as having infinite conductivity?
So...what is the equation in Meep for the "perfect conductor model" and how is Q calculated for a "perfect conductor"? -
#1529 by SeeShells on 09 Jan, 2016 20:46
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>> One would have to find out what was responsible for the numerical ill-conditioning...
Of course 11 Billion is not a real number by any means, but the real question I wanted answered is, can meep simulate a superconducting cavity with the current perfect metal settings or would we have to modify the drude model?
1) The problem is not just that 11 billion Q is not a real number, the problem is understanding why Meep gave that output. The problem is understanding why those Meep runs gave a negative huge value of Q of 11 billion, which means that energy was being produced in the Meep run at an enormous rate. If that bug is not understood (garbage into Meep = garbage out of Meep) then one cannot really have confidence in any other Meep run whether at ambient temperature or superconducting. It is a numerical problem with the model.
2) Meep is capable of modeling the superconducting regime, if the material properties are input correctly and if the Finite DIfference space mesh is converged and the finite difference time step is small enough, and the matrix is not ill-conditioned etc.
3) Given aero's information HarmInv was involved. It may be due to numerical ill-conditioning. One would have to find out what was responsible for the numerical ill-conditioning.
Correct.
Is there any reason we cannot do this, and do a cavity to run simulating superconducting materials?
Shell -
#1530 by X_RaY on 09 Jan, 2016 20:53
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No! But, why you like to explore superconducting cavity resonators until we are done with the normal conductors like copper?
>> One would have to find out what was responsible for the numerical ill-conditioning...
Of course 11 Billion is not a real number by any means, but the real question I wanted answered is, can meep simulate a superconducting cavity with the current perfect metal settings or would we have to modify the drude model?
1) The problem is not just that 11 billion Q is not a real number, the problem is understanding why Meep gave that output. The problem is understanding why those Meep runs gave a negative huge value of Q of 11 billion, which means that energy was being produced in the Meep run at an enormous rate. If that bug is not understood (garbage into Meep = garbage out of Meep) then one cannot really have confidence in any other Meep run whether at ambient temperature or superconducting. It is a numerical problem with the model.
2) Meep is capable of modeling the superconducting regime, if the material properties are input correctly and if the Finite DIfference space mesh is converged and the finite difference time step is small enough, and the matrix is not ill-conditioned etc.
3) Given aero's information HarmInv was involved. It may be due to numerical ill-conditioning. One would have to find out what was respowsible for the numerical ill-conditioning.
Correct.
Is there any reason we cannot do this, and do a cavity to run simulating superconducting materials?
Shell
As far as Ï know you are right now on the way to do this first with real experiments using a copper frustum at ambient temperature.
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#1531 by SeeShells on 09 Jan, 2016 21:01
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No! But, why you like to explore superconducting cavity resonators until we are done with the normal conductors like copper?
>> One would have to find out what was responsible for the numerical ill-conditioning...
Of course 11 Billion is not a real number by any means, but the real question I wanted answered is, can meep simulate a superconducting cavity with the current perfect metal settings or would we have to modify the drude model?
1) The problem is not just that 11 billion Q is not a real number, the problem is understanding why Meep gave that output. The problem is understanding why those Meep runs gave a negative huge value of Q of 11 billion, which means that energy was being produced in the Meep run at an enormous rate. If that bug is not understood (garbage into Meep = garbage out of Meep) then one cannot really have confidence in any other Meep run whether at ambient temperature or superconducting. It is a numerical problem with the model.
2) Meep is capable of modeling the superconducting regime, if the material properties are input correctly and if the Finite DIfference space mesh is converged and the finite difference time step is small enough, and the matrix is not ill-conditioned etc.
3) Given aero's information HarmInv was involved. It may be due to numerical ill-conditioning. One would have to find out what was respowsible for the numerical ill-conditioning.
Correct.
Is there any reason we cannot do this, and do a cavity to run simulating superconducting materials?
Shell
Two reasons.
One, we are still seeing a negative numbers being generated by meep and I believe we need to know why and what causes it.
Second, there is much discussion on the superconducting aspects of a EMDrive.
Last, if what I saw in the perfect conductors in runs in meep vs the Drude Cu models it is important to know that this maybe a area in the future to head for in designs.
Shell -
#1532 by aero on 09 Jan, 2016 21:50
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For information, if anyone wants to learn about Harminv and how it works, see attached.
And here is the Harminv Programmer's take on negative Q.
https://www.mail-archive.com/meep-discuss%40ab-initio.mit.edu/msg00161.html
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#1533 by Rodal on 09 Jan, 2016 22:39
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For information, if anyone wants to learn about Harminv and how it works, see attached.
And here is the Harminv Programmer's take on negative Q.
https://www.mail-archive.com/meep-discuss%40ab-initio.mit.edu/msg00161.html
one cause of a badly calculated Q: Meep simulations have been too short (0.01 microseconds), the Meep discuss page says:Quoterun the simulation longer
Running Meep for too short a time leads to malformed modes and travelling waves that may not be anywhere close to what steady state looks like, and incorrect Q calculations
QUESTION: have the cases for which Q was negative, been run for twice as long or longer?, if yes, did that bring Q into a more reasonable value ?
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#1534 by aero on 09 Jan, 2016 22:49
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For information, if anyone wants to learn about Harminv and how it works, see attached.
And here is the Harminv Programmer's take on negative Q.
https://www.mail-archive.com/meep-discuss%40ab-initio.mit.edu/msg00161.html
one cause of a badly calculated Q: Meep simulations have been too short (0.01 microseconds), the Meep discuss page says:Quoterun the simulation longer
Running Meep for too short a time leads to malformed modes and travelling waves that may not be anywhere close to what steady state looks like, and incorrect Q calculations
QUESTION: have the cases for which Q was negative, been run for twice as long or longer?, if yes, did that bring Q into a more reasonable value ?
As far as I can tell, running the simulation longer is the same as narrowing the bandwidth. There isn't a direct run length control as I use-
(run-sources+ (* gc T_meep)
(after-sources (harminv Ex (vector3 dlx dly dlz) fmeep BW 5))
-for Harminv runs ... Unless he is referring to the "idle
time" (* gc T_meep) between source cut-off and Harminv start. Narrowing the bandwidth gets very expensive very quickly in terms of run time but increasing the "idle time" is no big deal.
Maybe Meeper,ThereIWas3, had some success doing this, at least he did mention something along these lines. -
#1535 by zen-in on 09 Jan, 2016 23:47
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This may not be the correct method but you can try using a very high number for conductivity - like 1010. I used an estimate based on IC, the critical current for the superconductor (given the conditions) for some calculations I did a few years ago, since using infinity is the same thing as dividing by zero.It is definitely not that simple. See my post above.
One would have to look exactly as to:
A superconductor can lose it's state if external magnetic fields slowly oscillating fields are too strong or if photon energies of incident radiation is larger than the cooper binding energies.
If you manage to take this effects into account than the perfect metal approximation would be good.
1) How was the Q calculated by Meep. What equation was used to calculate Q?
2) What is meant by a "perfect conductor" model in Meep? What are the exact equations used inside it to model a "perfect conductor" ? Obviously if it was "perfect", conductivity would be infinite and the Q would be infinite. So if the Q is finite, does that mean that the perfect conductor model, is not so perfect, and the "perfect conductivity" has a finite value ?
What is the sense of calculating a numerical value for Q for a perfect metal if the perfect metal is modeled as having infinite conductivity?
So...what is the equation in Meep for the "perfect conductor model" and how is Q calculated for a "perfect conductor"? -
#1536 by aero on 10 Jan, 2016 00:56
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"Perfect metal" is implemented in Meep as a very large negative epsilon.
Q = - Re ω /2 Im ω by definition in Harminv. -
#1537 by rfcavity on 10 Jan, 2016 00:57
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In softwares I use, if you have a resonator and everything is ideal material (i.e. no loss) floating point errors can add a very tiny amount of extra energy which quickly grows without bound due to no energy sinks in the simulation. For a superconductor it seems to me you'd have to find what limits the energy storage of the cavity before it stops being a superconductor and I doubt meep or any other EM-only sim will have this.
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#1538 by rq3 on 10 Jan, 2016 01:05
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For information, if anyone wants to learn about Harminv and how it works, see attached.
And here is the Harminv Programmer's take on negative Q.
https://www.mail-archive.com/meep-discuss%40ab-initio.mit.edu/msg00161.html
one cause of a badly calculated Q: Meep simulations have been too short (0.01 microseconds), the Meep discuss page says:Quoterun the simulation longer
Running Meep for too short a time leads to malformed modes and travelling waves that may not be anywhere close to what steady state looks like, and incorrect Q calculations
QUESTION: have the cases for which Q was negative, been run for twice as long or longer?, if yes, did that bring Q into a more reasonable value ?
As far as I can tell, running the simulation longer is the same as narrowing the bandwidth. There isn't a direct run length control as I use-
(run-sources+ (* gc T_meep)
(after-sources (harminv Ex (vector3 dlx dly dlz) fmeep BW 5))
-for Harminv runs ... Unless he is referring to the "idle
time" (* gc T_meep) between source cut-off and Harminv start. Narrowing the bandwidth gets very expensive very quickly in terms of run time but increasing the "idle time" is no big deal.
Maybe Meeper,ThereIWas3, had some success doing this, at least he did mention something along these lines.
Mathematically narrowing the bandwidth is physically identical to increasing the Q. -
#1539 by ThinkerX on 10 Jan, 2016 01:41
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Quote
As I understand it, the objection here is that constant thrust at the levels claimed for the EM Drive lead directly to a violation of Conservation of Energy. However, suppose the thrust is NOT constant, but instead 'crashes' before reaching that point?
I am thinking here of a 'longer cycle.' Something on the order of 30-40 seconds of high 'thrust,' followed by an unavoidable 'crash' of at least several minutes with no thrust, yet still drawing power, then another 30-40 seconds of 'thrust,' followed by another 'crash.' A 'charge / discharge' cycle. To maintain 'constant acceleration,' you'd need several (10? 20?) EM Drives working in a timed sequence, and a corresponding increase in required power.
Does this represent a valid workaround for Conservation of Energy?QuoteSo, say you start at .1N/kW=100000µN/kW (where EM drive would start to be competitive with ion thruster, given contemporary flight ready sources of power -i.e. not advanced fission or fusion...-) and this is only valid for 1/10th of the time and during 9/10th each thruster still has to consume same nominal power (as per your hypothesis "yet still drawing power", which is hardly motivated : why not just switch off when 'crash' occurs and let things cool down to same state as that allowing a first pulse ?) then that gets us .01N/kW=10000µN/kW overall for the whole system (and added mass to the system, but total mass of EMdriven spacecraft doesn't play any role in the CoE issue). 10000µN/kW is not really interesting compared with ion thruster (given contemporary flight ready sources of power...), and would make power generation from apparent CoE breaking (breakeven) impractical (critical velocity>100km/s) , but still possible in principle, and that's obviously a problem with known physics.
I should have been clearer. The 'crash' (recharge with no apparent thrust) is mandatory. This is where the power for the active 'thrust generating' part of the cycle comes from. A sort of charge/discharge or capacitor effect. Without that long recharge/crash at one point, you don't get the 'thrust,' for a much shorter interval later on, even though you are still putting power into the frustum.
What brings this on is I note that while there are multiple experiments showing 'thrust,' none have lasted for more than a minute or so. In several cases - most recently with Seashell's, the thrust 'crashes' after a short while. Currently, when thought of at all, it is in the context of a 'bug.' I suspect it is a built in, unavoidable feature.
Likewise, with the MEEP simulations, Rodal and you both noted the effect goes exponential, but that we are seeing only the first 1/1000th (at the very most) of a much longer sequence. My suspicion is this exponential increase would eventually reverse itself.
Ω•m